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Lời giải:
a) ĐK: \(x\geq 0\)
\(4\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
\(\Leftrightarrow 4\sqrt{x}-2\sqrt{9}.\sqrt{x}+\sqrt{16}.\sqrt{x}=5\)
\(\Leftrightarrow 4\sqrt{x}-6\sqrt{x}+4\sqrt{x}=5\)
\(\Leftrightarrow 2\sqrt{x}=5\Rightarrow \sqrt{x}=\frac{5}{2}\Rightarrow x=\frac{25}{4}\) (thỏa man)
b) ĐK: \(x\geq -5\)
PT \(\Leftrightarrow \sqrt{4}.\sqrt{x+5}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9}.\sqrt{x+5}=6\)
\(\Leftrightarrow 2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow 3\sqrt{x+5}=6\Rightarrow \sqrt{x+5}=2\)
\(\Rightarrow x+5=2^2=4\Rightarrow x=-1\) (thỏa mãn)
ĐK: \(x\ge0\)\(4\sqrt{x}-2\sqrt{9x}+16\sqrt{x}=5\) 5 (=) \(\sqrt{x}\left(4-2\sqrt{9}+16\right)=5\) (=) \(\sqrt{x}.14=5\)(=) x=\(\frac{25}{196}\)
ĐK: \(x\ge-5\)PT(=) \(\sqrt{5+x}\left(\sqrt{4}-3+\frac{4}{3}.3\right)=6\) (=) \(\sqrt{5+x}.3=6\) (=)\(\sqrt{5+x}=2\)(=) X = -1 (nhận)
a) \(\sqrt{\left(2x-1\right)^2}=3\)
⇔ \(\left|2x-1\right|=3\)
⇔ \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
⇔ \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
b) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
ĐKXĐ : \(x\ge0\)
⇔ \(3\sqrt{x}-2\sqrt{3^2x}+\sqrt{4^2x}=5\)
⇔ \(3\sqrt{x}-2\cdot3\sqrt{x}+4\sqrt{x}=5\)
⇔ \(7\sqrt{x}-6\sqrt{x}=5\)
⇔ \(\sqrt{x}=5\)
⇔ \(x=25\)( tm )
c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)
ĐKXĐ : \(x\ge-5\)
⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)
⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot3\sqrt{x+5}=6\)
⇔ \(-\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)
⇔ \(\frac{5}{4}\sqrt{x+5}=6\)
⇔ \(\sqrt{x+5}=\frac{24}{5}\)
⇔ \(x+5=\frac{576}{25}\)
⇔ \(x=\frac{451}{25}\left(tm\right)\)
a) \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\) (ĐKXĐ : \(x\ge2\) )
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}-4\sqrt{x+3}-3\sqrt{x-2}=2\)
\(\Leftrightarrow\sqrt{x+3}=2\)
\(\Leftrightarrow x+3=4\)
\(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )
c) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\) (ĐKXĐ : \(x\ge-5\) )
\(\Leftrightarrow2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5}=4\)
\(\Leftrightarrow2\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\) ( Thỏa mãn ĐKXĐ )
Vậy.......
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ 2\sqrt{x-5}=4\\ \sqrt{x-5}=2\\ x-5=4\\ x=9\)
ĐK:x\(\ge5\)
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
Vậy S={9}
A, đk tự tìm
\(\sqrt{x^2+4x+3}=x-2\)
\(\Leftrightarrow x^2+4x+3-x^2+4x-4=0\)
\(\Leftrightarrow8x-1=0\)
\(\Leftrightarrow x=\frac{1}{8}\)
B, đk tự tìm
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9\left(x+5\right)}\)=6
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}\left(2-3+4\right)=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)
\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)
\(\Leftrightarrow9x-\left|7x+5\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow x=6\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)
b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x-5}=4-\sqrt{x+5}\)
\(\Leftrightarrow x-5=\left(4-\sqrt{x+5}\right)^2\)
\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)
\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)
\(\Leftrightarrow-5=21-8\sqrt{x+5}\)
\(\Leftrightarrow8\sqrt{x+5}=21+5\)
\(\Leftrightarrow8\sqrt{x+5}=26\)
\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)
\(\Leftrightarrow x+5=\dfrac{169}{16}\)
\(\Leftrightarrow x=\dfrac{169}{16}-5\)
\(\Leftrightarrow x=\dfrac{89}{16}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)
Nick cũ không đi giải lấy nick mới giải làm gì vậy Tuấn Anh Phan Nguyễn ? :D
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
ĐK:5+x\(\ge0\)\(\Leftrightarrow x\ge-5\)
\(\sqrt{4x+20}-3\sqrt{5+x}=6-\dfrac{4}{3}\sqrt{9x+45}\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}=6-\dfrac{4}{3}\sqrt{9\left(x+5\right)}\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}.3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)(tm)
đk: x≥-5
pt <=> \(2\sqrt{x+5}-3\sqrt{x+5}=6-\dfrac{4}{3}\cdot3\sqrt{x+5}\)
\(\Leftrightarrow-\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\Leftrightarrow\sqrt{x+5}=2\Leftrightarrow x+5=4\Leftrightarrow x=-1\left(tm\right)\)
vật pt có 1 nghiệm x=-1